Problem: Which of the following numbers is a multiple of 3? ${43,59,105,110,118}$
Solution: The multiples of $3$ are $3$ $6$ $9$ $12$ ..... In general, any number that leaves no remainder when divided by $3$ is considered a multiple of $3$ We can start by dividing each of our answer choices by $3$ $43 \div 3 = 14\text{ R }1$ $59 \div 3 = 19\text{ R }2$ $105 \div 3 = 35$ $110 \div 3 = 36\text{ R }2$ $118 \div 3 = 39\text{ R }1$ The only answer choice that leaves no remainder after the division is $105$ $ 35$ $3$ $105$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $3$ are contained within the prime factors of $105$ $105 = 3\times5\times7 3 = 3$ Therefore the only multiple of $3$ out of our choices is $105$. We can say that $105$ is divisible by $3$.